Question

differentiate 6/ root x​

Answer 1

Explanation:

answer is here ........

d/dx(6/√x)= 6√x/2.

=3√x.

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Answer 2

Answer:

Required value is $\frac{3}{ \sqrt{x} }$

Explanation:

Given,6/ root x

$= \frac{6}{ \sqrt{x} } \\ = 6 \times {x}^{ \frac{1}{2} }$

We want to differentiate it.

We know,

$\frac{d}{dx} ( {x}^{n} ) = n {x}^{n - 1}$

For example,

$\frac{d}{dx} ( {x}^{5} ) = 5 {x}^{5 - 1} \\ = 5 {x}^{4}$

Here given term is $6 {x}^{ \frac{1}{2} }$

Let $y = 6 {x}^{ \frac{1}{2} }$

We are differentiating both side with respect to y.

$\frac{dy}{dx} = \frac{d}{dx} ( 6{x}^{ \frac{1}{2} } ) \\ = 6 \frac{d}{dx} ( {x}^{ \frac{1}{2} } ) \\ = 6 \times \frac{1}{2} {x}^{ \frac{1}{2} - 1} \\ = 3 {x}^{ ( - \frac{1}{2}) } \\ = \frac{3}{ {x}^{ \frac{1}{2} } } \\ = \frac{3}{ \sqrt{x} }$

Some important derivatives formula,

$1. \frac{d}{dx} ( \sin(x) ) = \cos(x ) \\ 2. \frac{d}{dx} ( \cos(x) ) = - \sin(x) \\ 3. \frac{d}{dx} ( \tan(x) ) = {sec}^{2} x \\ 4. \frac{d}{dx} ( \cot(x) ) = - {cosec}^{2} x \\ 5. \frac{d}{dx} ( \sec(x) ) = \sec(x) \tan(x) \\ 6. \frac{d}{dx} (cosec(x)) = - cosecxcotx$

Know more about derivative,

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